Is There a Root for cos(x) = x in the Interval (0,1)?

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The discussion centers on whether the equation cos(x) = x has a root in the interval (0,1). Using the Intermediate Value Theorem (IVT), it is established that since f(0) = 1 and f(1) < 0, there must be at least one point c in (0,1) where f(c) = 0, indicating a solution exists. The continuity of the function f(x) = cos(x) - x on the closed interval [0,1] is also emphasized as crucial for applying the IVT. Participants express some confusion over terminology, suggesting "solution" may be a clearer term than "root." Ultimately, the IVT confirms the existence of a solution to the equation in the specified interval.
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From the Intermediate Value Theorem, is it guaranteed that there is a root of the given equation in the given interval?

cos(x)=x, (0,1)

cos(0)= 1
cox(1)= 0.540...

So using intermediate value theorem, no.

and

x=0 can't be possible because 0 was excluded in the domain by the round bracket so 0<x<1. Therefore there can't be a root. In other words the domain makes it so the graph is discontinuous at 0 (f(0) does not exist) and if the graph/function doesn't at that point there cannot be a root there.

The answer is yes, but I am struggling to understand why and I am not sure if the books answer is wrong or if I'm wrong.

*I also don't know why cos(x) = x , because when x=1 they aren't equal and would only be true at 1 point in this domain (when x=~0.79).
 
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It's probably better to examine the equivalent question; Is there a solution to ##\cos x-x=0## on the interval ##(0,1)##?

The Intermediate Value Theorem says
If ##f## is continuous on ##[a,b]## and ##L## is strictly between ##f(a)## and ##f(b)##, then there is at least one point ##c## in ##(a,b)## satisfying ##f(c)=L##.

If we put ##f(x)=\cos x-x##, your (revised) question is asking whether there is ##c## in ##(0,1)## satisfying ##f(c)=0##, right?

Given that ##f(0)=\cos 0-0=1>0## and ##f(1)=\cos 1-1<0## (you should check that), do you see how the IVT guarantees a ##c## in ##(0,1)## with ##f(c)=0##? Do you see how this is the same as saying ##\cos c=c##?

Note that you also need to say something regarding the continuity of ##f(x)=\cos x-x## on the closed interval ##[0,1]##.
 
Oh ok, that makes sense, I thought the question was asking me to do something else.

gopher_p said:
For what it's worth, I'm not a big fan of the use of the word "root" in the problem statement. It'd probably be better to use the word "solution" in that context.

Yeah, that would have made it much clearer imo.
 
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For what it's worth, I'm not a big fan of the use of the word "root" in the problem statement. It'd probably be better to use the word "solution" in that context.
 
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